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I understand how the capacitor "holds" charge. If one plate is positive, and the other is negative, the particles are attracted to each other. Connect both sides by a wire and instant flow.

However, I am confused why the charge gets built up on the capacitor in the first place when connected to a battery? There's no complete circuit.

However, everything I find online says that plate that's connected to the positive side of the battery will give up electrons that will flow through the battery to the other side? How does this even make sense?

As far as I understand it, a chemical reaction in the battery pushes the negative ions to one side and the positive ones to the other. Thus when I hook up the battery via a wire, the electrons have a direct path from the negative side to the positive side.

That's totally different than taking electrons from one side of a capacitor? To sum up, why is it that just nothing happens when I hook up a capacitor?

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Initially, the capacitor can be thought of as neutral. Meaning, it has no net charge. A mix of positive and negative charges on both plates (well call them upper and lower plates to identify which is which).

When you connect a battery, you have current flow. It doesn't matter if the circuit is open or close (from the perspective of the source, it does not know what is 1nm, 1mm, or 1m ahead of it). So you have these positive charges that start to loiter around the upper plate. At first, first, its 1 charge, 2 charge, 10 charges and it keeps doing that until the entire surface area of the plate is the same charge. At this point, there is no current flow on the capacitor because there is no longer a change in the electric field.

But as the feild is changing, (such as when the charges are accumulating), the positive charges on the upper plate, attract negative charges on the lower plate. But in order of a negative charge to accumulate on the lower plate, a positive charge on the lower plate needs to be....displaced...from the lower plate. Essentially being kicked out from that spot.The plates have finite space, so there is only enough room for so many charges. This repeats until every positive charge has attracted a negative charge.

This is how current flows in a capacitor. Through what is called, displacement current.

This is why current in an uncharged capacitor is instantaneous, and exponentially tapers off, as the charge builds up, there is less and less negative charges needed and so less and less positive charges get displaced.

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  • \$\begingroup\$ Perhaps this naive, but I was always under the impression that current can flow ONLY if there is a closed loop such as a circuit. With enough voltage, there's enough Joules/Coulomb to make that circuit through the air. However, there is STILL a circuit. Why is there a bit of current flow to begin with in your example? \$\endgroup\$ – user1357015 Oct 1 '15 at 2:41
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    \$\begingroup\$ There are two models in which engineers look at current. The circuit view, which is what is typically taught in most classes. Current flows like this, it does this, it does that. That model is used to analyze, and design circuits. However, that is not how current flows and board designers (those who do PCB layout) have to think about how current really flows when they do their layout (properly). Once you grasp the concept the displacement, you'll go back to terms like "charging the capacitor", " capacitor blocks dc", "capacitor passes AC" etc.. and they work for that model. They work good. \$\endgroup\$ – efox29 Oct 1 '15 at 2:47
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    \$\begingroup\$ @user1357015 So when you design on paper, you view current as one way. When you design or are interested in what current is actually doing (and not the effect it has), then you use another model. If you ever take an electromagnetics class, they cover Maxwell's Theories, and that guy knows how current flows. His 6 equations explain everything. \$\endgroup\$ – efox29 Oct 1 '15 at 2:50
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    \$\begingroup\$ @user1357015 see electronics.stackexchange.com/questions/172303/… \$\endgroup\$ – efox29 Oct 1 '15 at 2:52
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    \$\begingroup\$ @user1357015, there is a closed circuit. When the capacitor is charging there is a displacement current through the capacitor to complete the circuit. \$\endgroup\$ – The Photon Oct 1 '15 at 3:04
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This is hard to comprehend but there is a current in a capacitor. It is called a displacement current. It was demonstrated by Maxwell in his equations. This displacement current is not a direct flow of electrons but instead an electric field. A capacitor has an electric field between the plates that is the proverbial current flow.

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    \$\begingroup\$ Displacement current isn't an electric field, but a change in electric field. Once the capacitor is charged, the field is constant and there's no displacement current. While the capacitor is charging, the field is changing, and there is an associated displacement current. \$\endgroup\$ – The Photon Oct 1 '15 at 3:03
  • \$\begingroup\$ @ThePhoton Agreed, I was paraphrasing. \$\endgroup\$ – vini_i Oct 1 '15 at 9:49
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You might be getting into trouble by visualizing a capacitor as being like two separate plates. Are you assuming that you can remove 1uC of charge from one plate, but not removing any charge from the other? That may be OK for physics homework (where capacitors are two metal spheres at great distance.) But in electronics, capacitors are relatively huge plates, spaced microscopically apart. An electrical engineer's capacitor isn't like two separate metal spheres. Instead it behaves like a single ball with a microscopic gap sliced all the way through it. Two solid polished hemispheres separated by a micron of plastic film.

         ____  ____
      __/    ||    \__
     /       ||       \
   /         ||         \       "Engineer's Capacitor"
  |          ||          |        
 |           ||           |     A spilt metal sphere with
|            ||            |    a very narrow gap between 
|            ||            |    the two halves
|            ||            |
 |           ||           |
  |          ||          |
   \         ||         /
     \__     ||     __/
        \____||____/

In other words, the capacitance to distant ground is tiny, while capacitance between the two hemispheres is enormous. The main effect is that, if you try to force some charge continuously into one plate of this "engineer's capacitor," that charge instantly spreads to the outer surface of both plates. And then, a large voltage appears on both plates WRT ground, halting the current. The only way to avoid this effect, and to create a continuing current, is by treating the capacitor as a two-terminal conductor. In that case we cannot push charge into the first plate unless simultaneously we remove it from the second.

The capacitor only works right if we pretend that it's some sort of wire. With wires, if we try to push in more charge than we remove, both ends of the wire immediately charge up to fantastic values of voltage, and this blocks any further current. A capacitor in electronics does the same thing. More: the engineer's capacitor

Why does this occur? I personally find this fascinating. If we dump positive charge into one hemisphere, it spreads out over the surface, including that part of the surface down in the gap between hemispheres. The positive charge that went into the gap attracts equal negative charge to the other side of the gap. But this leaves a positive excess in the second hemisphere, and this excess self-repels, traveling to the outer surface of the second hemisphere. If the surface excess on the second half grows too large, then the opposite happens, and excess negative in the gap will create excess negative on the outside of the original hemisphere. When things have settled down, half the charge that we placed on one hemisphere has seemingly traveled to the outer surface of the second hemisphere! And, down in the gap, each flat face also has half the amount of deposited charge, but on the second hemisphere it's opposite polarity.

So, capacitors act like single metal objects, where charge seemingly leaps the gap with ease. Electrically, the two halves are coupled together very, very tightly. To produce a current, we're required to connect them in a circuit, where the path for current is through the component, and back out again. And then of course the capacitor equation takes hold ...for every second the current through the capacitor exists, the voltage between the plates grows in proportion to I/C, and energy is stored in the gap. A capacitor is a 2-terminal component, and not anything like an open circuit or a pair of widely-spaced plates. The path for current is through.

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